Chapter 9: StoichiometryChapter 9: Stoichiometry

9.1 Mole to Mole 9.1 Mole to Mole

Objective: To perform mole to Objective: To perform mole to mole conversion problems.mole conversion problems.

StoichiometryStoichiometry

StoichiometryStoichiometry is the branch of is the branch of chemistry that deals with quantities chemistry that deals with quantities of substances in chemical reactions.of substances in chemical reactions.

Mole to Mole Problems: The Mole to Mole Problems: The StepsSteps

1. Write chemical equation1. Write chemical equation

2. Balance chemical equation using coefficients2. Balance chemical equation using coefficients

3. Use the following setup to perform 3. Use the following setup to perform calculation:calculation:

A is the A is the known known quantityquantity

B is the B is the unknown quantityunknown quantity

4. Don’t forget units on your final answer!!4. Don’t forget units on your final answer!!

molBeqnmolA

eqnmolBmolA ___

.)(___

.)(______

Mole Ratio

9.2 Mole to Mass and Mass to 9.2 Mole to Mass and Mass to MoleMole

Objective: To perform mole to mass Objective: To perform mole to mass and mass to mole conversion and mass to mole conversion problems.problems.

Mole to MassMole to Mass

Moles Given

Given Moles (Eqn.)

Unknown Moles (Eqn.)

Molar Mass Unknown1 mole Unknown

Grams Unknown

Mole Ratio

Mass to MoleMass to Mole

Grams Given

Molar Mass Given

Given Moles (Eqn.)

1 mol GivenUnknown Moles (Eqn.) Moles

Unknown

Mole Ratio

9.3 Mass to Mass9.3 Mass to Mass

Objective: To perform mass to mass Objective: To perform mass to mass conversion problems.conversion problems.

Stoichiometry RoadmapStoichiometry Roadmap

Grams Given

Molar Mass Given

Given Moles (Eqn.)

1 mol GivenUnknown Moles (Eqn.)

Molar Mass Unknown1 mole Unknown

Grams Unknown Mole Ratio

9.4 Limiting Reactant9.4 Limiting Reactant

Objectives: Objectives:

(1)(1)To calculate the To calculate the theoretical yield theoretical yield of a chemical reactions.of a chemical reactions.

(2)(2)To determine the To determine the limiting reactant limiting reactant and and excess reactantexcess reactant in a chemical in a chemical reaction.reaction.

Limiting ReactantLimiting Reactant

Any reactant that is used up first in a Any reactant that is used up first in a chemical reaction.chemical reaction.

It determines the amount of product It determines the amount of product that can be formed in the reaction.that can be formed in the reaction.

Excess ReactantExcess Reactant

The reactant that is not completely The reactant that is not completely used up in a reaction.used up in a reaction.

Limiting Reactant ProblemsLimiting Reactant Problems

Use the mass to Use the mass to mass conversionsmass conversions

Grams Given 1 mol Given

Unknown Moles

Molar Mass of Unknown

 Molar Mass

Given Given Moles 1 mol of Unknown

ExampleExample

Copper reacts with sulfur to form Copper reacts with sulfur to form copper(I) sulfide according to the copper(I) sulfide according to the following balanced equation:following balanced equation:

2Cu + S 2Cu + S Cu Cu22SS

What is the limiting reactant when 80.0 What is the limiting reactant when 80.0 grams of Cu reacts with 25.0 grams of grams of Cu reacts with 25.0 grams of S?S?

ExampleExample2Cu + S 2Cu + S Cu Cu22SS

The general equation is:The general equation is:Grams Given 1 mol Given

Unknown Moles

Molar Mass of Unknown

 Molar Mass

Given Given Moles 1 mol of Unknown

80.0 g Cu 1 mol Cu

1 mol Cu2S

159.17 g Cu2S

  63.55 g Cu 2 mol Cu 1 mol of Cu2S

25.0 g S 1 mol S

1 mol Cu2S 159.17 g Cu2S

  32.07 g S 1 mol S 1 mol of Cu2S

Start with Copper:

Now use Sulfur:

= 100.19 g Cu2S

= 124.08 g Cu2S

ExampleExample

The limiting reactant is copper.The limiting reactant is copper.

The excess reactant is sulfur.The excess reactant is sulfur.

The amount of CuThe amount of Cu22S that is produced S that is produced is 100.19 g Cuis 100.19 g Cu22S. S.

9.5 Percent Yield9.5 Percent Yield

Objective: To calculate Objective: To calculate percent percent yieldyield..

Introduction to Percent Introduction to Percent Yield…Yield…

If you get 15 out of 20 questions correct on If you get 15 out of 20 questions correct on a test, what percentage did you receive on a test, what percentage did you receive on the test? How did you figure this out?the test? How did you figure this out?

15/20 x 100 = 15/20 x 100 = 75%75%

If Sammy Sosa gets 25 hits in the month of If Sammy Sosa gets 25 hits in the month of May and has 113 bats, what is his batting May and has 113 bats, what is his batting average for the month of May? Explain average for the month of May? Explain how you arrived at your answer.how you arrived at your answer.

25/113 = 25/113 = 0.2210.221As a percentage, this is written, 25/113 x As a percentage, this is written, 25/113 x

100 = 100 = 22.1%22.1%

Percent YieldPercent Yield

··Percent yield is the ratio of the actual yield to Percent yield is the ratio of the actual yield to the theoretical yield for a chemical reaction the theoretical yield for a chemical reaction expressed as a percentage.expressed as a percentage.

·· It is a measure of efficiency of a reaction. It is a measure of efficiency of a reaction.

Percent Yield = Percent Yield = actual yieldactual yield x x 100%100%

theoretical yieldtheoretical yield

Actual YieldActual Yield

The amount of product that forms The amount of product that forms when a reaction is carried out in the when a reaction is carried out in the laboratory.laboratory.

Theoretical YieldTheoretical Yield

The amount of product that could The amount of product that could form during a reaction calculated form during a reaction calculated from a balanced chemical equation.from a balanced chemical equation.

It represents the maximum amount It represents the maximum amount of product that could be formed from of product that could be formed from a given amount of reactant.a given amount of reactant.

Example #1Example #1The equation for the complete combustion of The equation for the complete combustion of

ethene (Cethene (C22HH22) is ) is

2C2C22HH22 + 5O + 5O22 4CO 4CO22 + 2H + 2H22OO

1.1. If 0.10 g of CIf 0.10 g of C22HH22 is reacted with 201.60 g of O is reacted with 201.60 g of O22, , identify the limiting reactant.identify the limiting reactant.

2.2. What is the theoretical yield of HWhat is the theoretical yield of H22O?O?

3.3. If the actual yield of HIf the actual yield of H22O is 0.05 g, calculate O is 0.05 g, calculate the percent yield.the percent yield.

Example #1Example #12C2C22HH22 + 5O + 5O22 4CO 4CO22 + 2H + 2H22OO

0.10 g C2H2 1 mol C2H2 2 mol H2O 18.00 g H2O

  26.02 g C2H2 2 mol C2H2 1 mol H2O

201.60 g O2

1 mol O2

2 mol H2O18.00 g H2O

 

32.00 g O2

5 mol O2

1 mol H2O

Start with C2H2:

Now start with O2:

= 0.07 g H2O

= 45.36 g H2O

Theoretical YieldLimiting Reactant = C2H2

Example #1Example #12C2C22HH22 + 5O + 5O22 4CO 4CO22 + 2H + 2H22OO

Actual Yield = 0.05 g HActual Yield = 0.05 g H22OO

Theoretical Yield = 0.07 g HTheoretical Yield = 0.07 g H22OO

Percent Yield = actual yieldPercent Yield = actual yield x 100%x 100% theoretical yieldtheoretical yield

Percent Yield = 0.05 g HPercent Yield = 0.05 g H22O x 100% = O x 100% = 71.43 71.43 %%

0.07 g H0.07 g H22OO

Example #2Example #2

Determine the percent yield for the Determine the percent yield for the reaction between 3.74 g of Na and excess reaction between 3.74 g of Na and excess OO22 if 5.34 g of Na if 5.34 g of Na22OO22 is recovered? is recovered?

First, write the chemical equation:First, write the chemical equation:

Na + ONa + O22 Na Na22OO22

Second, balance the chemical equation:Second, balance the chemical equation:

2Na + O2Na + O22 Na Na22OO22

Example #2Example #22Na + O2Na + O22 Na Na22OO22

Solve the mass-mass problem, starting with Na:Solve the mass-mass problem, starting with Na:

Actual Yield = 5.34 g NaActual Yield = 5.34 g Na22OO22

Theoretical Yield = 6.34 g NaTheoretical Yield = 6.34 g Na22OO22

Percent Yield = Percent Yield = actual yieldactual yield x 100% x 100%theoretical yieldtheoretical yield

Percent Yield = Percent Yield = 5.34 g x 100% = 5.34 g x 100% = 84.23%84.23%6.34 g 6.34 g

3.74 g Na 1 mol Na

1 mol Na2O2 77.98 g Na2O2

22.99 g Na 2 mol Na 1 mol Na2O2

= 6.34 g Na2O2